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blsea [12.9K]
3 years ago
13

What equation would you use to find the time taken for an object to travel a set distance?

Physics
1 answer:
3241004551 [841]3 years ago
6 0

<em>Time = (the set distance) / (the object's traveling speed)</em>

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I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......
pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100cm^{2} x 10cm = 1000cm^{3}

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

\frac{1g}{cm^{3} } x 1000cm^{3} = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ \frac{8g}{cm^{3} } = 2.5 cm^{3}

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5cm^{3} overflowed. So now we the same process as in number a) just with a few adjustments.

\frac{1g}{cm^{3} } x (1000cm^{3} - 2.5cm^{3}) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0
3 years ago
. Imagine that you are standing at the center of a giant bowl of gelatin. What type of wave will you make across the top of the
vichka [17]
Transverse wave as the wave is going up and down no compressions
3 0
3 years ago
When a cup is placed on a table, which force prevents the cup from falling to the ground?
zhuklara [117]

Answer:

B. normal force

Explanation:

Because there is no frictional or resistance force. However gravitational force is applied downroad from the center of the cup thus the contact force that is perpendicular to the surface that an object contacts which is the normal force exerted upward from the table that prevents an object from falling.

6 0
3 years ago
If a wave had a wave speed of 1000 m/s and a frequency of 500 Hz, what is its wavelength?
Reil [10]

Answer:

<h2>2 meters</h2>

Explanation:

<h2>Wavelength = Speed/Frequency </h2><h2>1000 m/s ÷ 500 hz </h2><h2>2 m</h2><h2>hz = s</h2><h2>Hopes this helps. Mark as brainlest plz!</h2>

7 0
3 years ago
For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.
Kamila [148]

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

E = hv

E = \frac{hc}{\lambda}

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}

E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})

<u>E = 4.96 x 10³ eV</u>

<u></u>

B)

The energy of a particle at rest is given as:

E = m_{0}c^2

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (9.1 x 10^{-31} kg)(3  x  10^8 m/s)^2\\

E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\

<u>E = 4.19 x 10⁴ eV</u>

<u></u>

C)

The energy of a particle at rest is given as:

E = m_{0}c^2

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (6.64 x 10^{-27} kg)(3  x  10^8 m/s)^2\\

E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\

<u>E = 3.73 x 10⁹ eV</u>

8 0
2 years ago
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